Question

If 8ml of uncombined ${{O}_{2}}$ remain after exploding ${{O}_{2}}$ with 4ml of hydrogen, the volume in mL of ${{O}_{2}}$ originally was:
(A) 12
(B) 2
(C) 10
(D) 4

Answer 1

Hint: Start by writing the balanced chemical reaction. Accordingly equate with number of moles of hydrogen to that of oxygen using the reaction, to obtain the amount of oxygen used up in the reaction. Then, add that to remaining oxygen to get the answer.

Complete step by step answer:
-Let us first write the reaction taking place.

\[{{H}_{2}}+{}^{1}/{}_{2}{{O}_{2}}\to {{H}_{2}}O\]

-One mole of hydrogen gas reacts with half mole of oxygen gas to give one mole of water.
-According to the question, 4ml of hydrogen was used in this reaction and 8ml of unreacted oxygen was left behind.
-So, if 4ml of hydrogen was used in the reaction then half the volume of 4ml will be the oxygen used up in the reaction.
-Therefore, 2ml of oxygen will be used up in the reaction.
-Now, we need to calculate the initial volume of oxygen. By just adding the amount of oxygen used in the reaction and the volume of unreacted oxygen, we will get the initial volume of oxygen.
- Therefore, 2ml + 8ml = 10ml
-Hence, the volume of oxygen taken originally was 10ml.
So, the correct answer is “Option C”.

Note: Make sure you write the chemical reaction for 1 mole of hydrogen. In case, you use 2 moles of hydrogen to make balancing easier, then you will have to calculate the ratio of the number of moles of hydrogen and oxygen and then proceed. Even using that method, the answer will be the same.